Identifying Quadric Surfaces by Equation

[zhkhelpneeded]

Hi. I am reviewing for an upcoming multivariate calculus final.
I was looking at this video Link for extra practice.

My doubt is with equation 1: 2z^2 + 3y^2 - x = 1

By looking at it, I thought it was a hyperboloid with 1 sheet because that's a basic quadric surface that is set to a constant and has two negatives.
However, the person in the video said it's an elliptic paraboloid. But the basic equation for that surface is x^2 + y^2 = z. My question is how did they get id of the 1 in the original equation

Like they explained it using xy-traces, but I'm still confused. The xy-trace here would be 3y^2 - x = 1. The person said this is a parabola. How is this a parabola when the variables are different? Does that affect anything?

 

[Dr.Peterson]

However, the person in the video said it's an elliptic paraboloid. But the basic equation for that surface is x^2 + y^2 = z. My question is how did they get rid of the 1 in the original equation

How is this a parabola when the variables are different? Does that affect anything?

There are two issues here. (I have not watched the video.)

First, it doesn't matter where each variable appears. Your example of x^2 + y^2 = z is a circular paraboloid about the z-axis; 2z^2 + 3y^2 = x would be an elliptical paraboloid about the x-axis. It's the same basic shape, facing in a different direction.

Second, the constant indicates that there is a shift (translation). Your equation can be written as 2z^2 + 3y^2 = (x+1), where x has been replaced by (x+1), which shifts the graph in the negative x direction.

It's quite possible that the subject of the video is not quite the same as that of your example. Does the example come from the video or the course?

 

[JeffM]

The standard way to write a parabolic function is [imath]y = \alpha x^2 + \beta[/imath].

You probably learned that in first year algebra. But that standard form is based on the convention that univariate functions are defined using x as the independent variable and y as the dependent variable. But that convention is not a law of algebra. It is certainly allowed to write [imath]x = f(y).[/imath] If you do that, then the form would become [imath]x = \alpha y^2 + \beta[/imath].

The point that the video is making, or one of the main points, is that the naming conventions for quadric surfaces depends on what the majority of surfaces look like when holding different variables constant.

So holding z constant, we get

[math]z = \gamma \implies 2z^2 = 2 \gamma^2 \implies\\ 2z^2 + 3y^2 - x = 1 = 2 \gamma ^2 + y^2 + x = 1 \implies \\ x = \alpha y^2 + \beta, \text { where } \alpha = 3 \text { and } \beta = 2 \gamma^2 - 1.[/math]
That describes a parabola with x as the independent variable.

The same thing happens when we hold y constant.

So there are two parabolas, and our naming convention says that makes the surface a parabaloid.

If we hold x constant, we get

[math]2z^2 + 3y^2 = \alpha \ne 0 \implies \dfrac{z^2}{3\alpha} + \dfrac{y^2}{2 \alpha} = 1.[/math]
That describes an ellipse.

So an elliptic parabaloid.

EDIT: A question for you. I essentially implied that we were holding x constant at any value EXCEPT - 1. Why was that legitimate?